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Re: Superheat formula for charging capilary systems

No problem. It's a fairly simple exercise. Get out your superheat chart and look at a column near the center. We're going to assume that the column represents a given indoor wet bulb temp at various ambient temps. Note that for every 2 degree rise in ambient the target surperheat decreases by 1 degree. This is "approximately true". In fact its very close to being an exact relationship. So, for every 1 degree rise in ambient there will be a 0.5 degree drop in target superheat. This holds for every other column as well, approximately. So this seems to be a general rule that is "close enough for guvment work" as they say.

Now if you look at each row, representing a given ambient temp at various indoor wet bulb temps, then a similar linear relationship holds. In this case for every 2 degree rise in indoor wet bulb there's a 3 degree rise in target superheat. And again, this holds approximately for every row on the chart. The trick is now to arrange these proportional relationships in a single expression that will reflect these changes =. The standard point slope equation found in your elementary geometry book won't do. The equation that I derived is a variation of the point slope formula for a plane in 3D rather than a line in 2D. I've actually graphed this formula on my TI 89 Titanium in a 3-D rotatable form. It forms a 4 sided polygon with uneqal sides and angles, kind of shaped like a crooked wedge.

In any case, it was a bit of trial and error on my part, since I couldn't find an actual outline of how to convert these relationships into a single formula. I tried using the best fit program on my TI but ended up with a polynomial equation that was as lengthy as the superheat chart. No good. I tried again.

I needed the target superheat to drop 1 degree for every 2 degree rise in ambient, so I wrote down

Target SH = 0.5 ambient + C

( where C is an unknown constant)

Now it's easy to see that regardless of what ambient is, the target superheat is going to change by one half the amount that ambient changes when using that formula. Then similarly for iwb I needed the target superheat to go up 3 degrees for every 2 degree rise in iwb. So I figured I need a second term.

Target SH = 0.5 ambient + 1.5 iwb + C

Now its easy to see that at any given ambient and any given constant C that as iwb changes 2 degrees the Target SH is going to change 3 degrees. Now since both relationships are linear, the target superheat will vary by the correct amount regardless of whether ambient is held constant or not, so this has to be the correct formula. The only remaining task is to establish a value for C. Well it wasn't that simple. As it turned out I had to rearrange the formula like this to get the correct results.

Target SH = 1.5 iwb - 0.5 ambient + C

The ambient term had to be subtracted in order to get the target superheat to drop as ambient increased. The iwb term is added (positive) because the target superheat has to rise as iwb rises. The relationships are in reverse directions from one another.

Now look again at the superheat chart. You'll see that there are diagonals that contain the same target superheat all across them. From top left to bottom right. This means that if C is going to be constant, and if we take Target SH equal to say 10 deg then we'd have to have some arrangment of the two terms and C that would always give us the answer 10.

By selecting actual values for iwb and ambient along one of those diagonals I determined that the final form of the equation would have to be

Target SH = 1.5 iwb - 0.5 ambient - 40

The value 40 was actually a compromise that would allow the formula to remain fairly accurate across the board. The actual values required ranged I think from 38 to 42 or something along those lines, so I just averaged them and ended up using 40. That was accurate in the middle of the chart so I left it at that. I actually had 3 superheat charts in front of me when I was working this out. A Carrier, Rheem, and Trane chart. I wanted the formula to be a compromise that would fit all of those charts about equally as well. The differences between them amounted to a couple degrees at most in places.

This form (above) seemed a bit awckward to work in your head, so I simplified the factors, turning them into whole numbers, by multiplying through by 2/2 which gives

Target SH = [2(1.5)iwb - 2(0.5) ambient - 2(40)] /2

which of course reduces to

Target SH = (3 x iwb - ambient - 80) /2

I don't however use either of those forms of the formula in practice. I use this one instead

Target SH = ((iwb - 60) x 1.5) + (100 - ambient)/2

So I measure for instance an indoor wet bulb temp of 68. Very quickly I can calculate in my head 1.5 x 8 = 12. (The 8 comes from 68 minus 60. ) That gives me 12 SH for the indoor component. Then I might measure an ambient of 90. Very quickly in my head I get 1/2 of 10 = 5. So I have 5 degrees SH for the outdoor component. 12 + 5 is 17, so that's my Target SH.

With a little algebra you should be able to find that all three of these formulas are mathmetically equivalent to one another.

As per your PM request, my name is Richard Perry, no need to answer that via PM. I'm a service technician in central Arkansas. That's all your audience probably needs to know about the author of that formula. I've been a technician for the last 25 years in the same town.

BTW I actually ended up through logical argument with a formula more like the last one, and then derived the first one from it. There was algebra spread out over 3 or 4 pages that even I couldn't make sense of a few days later. My methods seem disorganized, but they sometimes work. after I had already derived the formula the hard way, I cracked open my College Algebra/Pre Calculus text and found out how to combine two point slope equations into one equation with two independent variables. I got the same equation a lot eaisier that way, but damned if I remember exaclty how I did that. I hate Calculus, that's why I bought a calculator that could do it for me.

Last edited by nike123; 01-04-2010 at 03:02 PM.

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